The Student Room Group

m1 q5 from may 2004

a boat of mass 400kg is held at rest on a slipway by a rope, the boat is modlled as a paricle on a rough plane inclined at 15 degees to the horizontal . The coefficient of friction is 0.2. the rope is modelled as a light, inextensible string, parallel to the line of greatest slop. The boat is in equilirium and on the point of slipping down the plane.

Calculate the Tension in the string?

the boat is 50m from the botom of the slipway. the rope is detached from the boat and the boat slides down the slipway.

Calculate the time taken for the boat to slide to the bottom of the slipway?

please help me :confused:
Reply 1
Draw a diagram for every situation in mechanics, it always helps.

Resolving perpendicular:
R=400gcos15

Resolving parallel:
T + 0.2R = 400gsin15
T = 400gsin15 - 0.2[400gcos15]
T = 400g[sin15-0.2cos15]
T = 257.3N

F=ma
400gsin15-0.2R = 400a
400gsin15 - 0.2[400gcos15] = 400a
a= 0.643ms^-2
s=50, a=0.643, t=0, t=?
s=ut+0.5at^2
50=(0.643/2)t^2
t=12.5sec(3sf)
ruth_lou
a boat of mass 400kg is held at rest on a slipway by a rope, the boat is modlled as a paricle on a rough plane inclined at 15 degees to the horizontal . The coefficient of friction is 0.2. the rope is modelled as a light, inextensible string, parallel to the line of greatest slop. The boat is in equilirium and on the point of slipping down the plane.

Calculate the Tension in the string?

the boat is 50m from the botom of the slipway. the rope is detached from the boat and the boat slides down the slipway.

Calculate the time taken for the boat to slide to the bottom of the slipway?

please help me :confused:

Tension
Resolving forces perpendicular to plane.
Forces in equilibrium so resultant force = 0

R - 400gcos15 = 0
R = 400gcos15 (1)

Resolving forces down the plane.
400gsin15 - F - T = 0 (F = mu*R so F = 0.2R)
400gsin15 - 0.2R - T = 0
T = 400gsin 15 - 0.2R (2)

Substutute 1 into 2.
T = 400gsin15 - 0.2(400gcos15)
T = 257.3N

Time taken
Once the boat is detached, there is no tension in the string and we get acceleration.

Therefore;
Forces down the plane.
400gsin15 - 0.2R = ma (F=ma)
400gsin15 - 0.2(400gcos15) = 400a
a = 257.3/400 = 0.643ms^-2

u = 0, s = 50m, a = 0.643ms^-2, t = ?
s = ut + 0.5at^2
50 = 0.322t^2
t = 12.5 seconds.
Reply 3
ah thank you i was half way there!